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Thread: Intergration by substitution

  1. #1
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    Intergration by substitution

    Do do you this one please
    Attached Thumbnails Attached Thumbnails Intergration by substitution-001.jpg   Intergration by substitution-002.jpg  
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle \int\frac{1+x}{1+x^2}dx=\int\frac{1}{1+x^2}dx+\int \frac{x}{1+x^2}dx=$
    $\displaystyle \displaystyle =\arctan x+\frac{1}{2}\int\frac{2x}{1+x^2}dx=\arctan x+\frac{1}{2}\ln (1+x^2)+C$
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Do do you this one please
    $\displaystyle \int \frac {1 + x}{1 + x^2}~dx = \int \frac {1}{1 + x^2}~dx + \int \frac {x}{1 + x^2}~dx$

    $\displaystyle = \arctan x + \int \frac {x}{1 + x^2}~dx$

    We proceed by substitution for the second integral:

    Let $\displaystyle u = 1 + x^2$

    $\displaystyle \Rightarrow du = 2x ~dx$

    $\displaystyle \Rightarrow \frac {1}{2} du = x ~dx$


    $\displaystyle \Rightarrow \int \frac {1 + x}{1 + x^2}~dx = \arctan x + \frac {1}{2} \int \frac {1}{u} ~du$

    $\displaystyle = \arctan x + \frac {1}{2} \ln u + C$

    $\displaystyle = \boxed { \arctan x + \frac {1}{2} \ln \left( 1 + x^2 \right) + C }$


    EDIT: You're too fast for me red_dog!
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  4. #4
    MHF Contributor red_dog's Avatar
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    $\displaystyle f(x)=\tan ^3x$ is an odd function and $\displaystyle \displaystyle \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$ is a symetric interval, so the integral is $\displaystyle 0$.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    I will do the indefinite integral for the second, i leave plugging in the limits to you

    $\displaystyle \int \tan^3 x ~dx = \int \tan^2 x \tan x ~dx$

    $\displaystyle = \int \left( \sec^2 x - 1 \right) \tan x ~dx$

    $\displaystyle = \int \left( \sec^2 x \tan x - \tan x \right)~dx$

    $\displaystyle = \int \sec^2 x \tan x ~dx - \int \tan x ~dx$

    $\displaystyle = \int \sec^2 x \tan x ~dx - \ln | \sec x | + C$

    We proceed by substitution for the first integral:

    Let $\displaystyle u = \tan x$

    $\displaystyle \Rightarrow du = \sec^2 x ~dx$

    $\displaystyle \Rightarrow \int \tan^3 x ~dx = \int u ~du - \ln | \sec x | + C$

    $\displaystyle = \frac {1}{2} u^2 - \ln | \sec x | + C$

    $\displaystyle = \boxed { \frac {1}{2} \tan^2 x - \ln | \sec x | + C }$
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    $\displaystyle f(x)=\tan ^3x$ is an odd function and $\displaystyle \displaystyle \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$ is a symetric interval, so the integral is $\displaystyle 0$.
    Ah, i never noticed that (never paid attention to the limits, i planned on finding the indefinate integral from the start). That's true, the integral is 0. However, if what you actually wanted was the area, you would find $\displaystyle 2 \int_{0}^{ \frac { \pi}{6}} \tan^3 x~dx$
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  7. #7
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    Another possible solution, similar to Jhevon's is:

    $\displaystyle \int{tan^{3}(x)}dx=\int{(sec^{2}(x)-1)tan(x)}dx=sec^{2}(x)tan(x)-tan(x)$

    $\displaystyle \int{sec^{2}(x)tan(x)}dx$

    Now, let $\displaystyle u=sec(x), \;\ du=sec(x)tan(x)dx$

    $\displaystyle \int{u}du=\frac{1}{2}u^{2}=\frac{1}{2}sec^{2}(x)$

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    $\displaystyle \int{tan(x)}dx$

    $\displaystyle \int\frac{sin(x)}{cos(x)}dx$

    Let $\displaystyle u=cos(x), \;\ -du=sin(x)dx$

    $\displaystyle -\int\frac{1}{u}du=ln(u)=-ln(cos(x))$

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    So, we have $\displaystyle \boxed{\frac{1}{2}sec^{2}(x)+ln(cos(x))+C}$


    $\displaystyle \frac{1}{2}tan^{2}(x)-ln(sec(x))+\frac{1}{2}=\frac{1}{2}sec^{2}(x)+ln(co s(x))$
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  8. #8
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    Quote Originally Posted by red_dog View Post
    $\displaystyle f(x)=\tan ^3x$ is an odd function and $\displaystyle \displaystyle \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$ is a symetric interval, so the integral is $\displaystyle 0$.
    This is the simplest method of all.
    The integral of an odd function on a symmetric interval is always zero
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